Question

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plz solv now..​

Answer 1

Question:

A constant force acts on an object of mass 5 kg for a duration of 2 s. It increases the object's velocity from 3 m/s to 7 m/s. Find the magnitude of the applied force. Now, if the force was applied for a duration of 5 s, what would be the final velocity of the object ?

Answer:

Force = 10 N

Final velocity = 13 m/s

Explanation:

Given:

• Mass of the body = 5 kg
• Time for which the force was applied = 2 s
• Initial velocity = 3 m/s
• Final velocity = 7 m/s

To Find:

• Velocity of  the body if the same force was applied for a time period of 5 s

Concept:

First find the accleration and substitute the data in Newton's second law of motion to get the value of force. Substituting the value of force, we get the value of final velocity.

Solution:

First we have to find the acceleration of the object.

Acceleration is given by,

a = (v - u)/t

where a = acceleration

           v = final velocity

           u = initial velocity

           t = time taken

Substituing the data,

a = (7 - 3)/2

a = 4/2

a = 2 m/s²

Hence the acceleration of the body is 2 m/s².

Now by Newton's second law we know that,

F =  ma

where f = force

          m = mass

           a = acceleration

Substituting the data,

F = 5 × 2

F = 10 N

Hence force acting on the body is 10 N.

Now in the second case,

Initial velocity = 3 m/s

Time taken = 5 s

Force applied = 10 N

Now by the equation,

F = m (v - u)/t

Substitute the data,

10 = 5 × (v - 3)/5

10 = v - 3

v = 10 + 3

v = 13 m/s

Hence the final velocity of the object is 13 m/s.